Mechanics of Materials
Chapter 4
Shear and Moment In Beams
4.1 Introduction
The termbeamrefers to a slender bar that carries transverse
loading; that is, the applied force are perpendicular to the bar.
In a beam, the internal force system consist of a shear force and
a bending momentacting on the cross section of the bar. The
shear force and the bending moment usually vary continuously
along the length of the beam.
The internal forces give rise to twokinds of stresses on a
transverse section of a beam: (1) normal stressthat is caused by
bending momentand (2) shear stressdue to the shear force.
Knowing thedistribution of the shear force and the bending
moment in a beam is essential forthe computation of stresses
and deformations. Which will be investigated in subsequent
chapters.
Figure 4.1 (a) Statically determinate beams.
4.2Supports and Loads
Beams are classified according to their supports. A simply
supported beam,shown in Fig. 4.1 (a). The pin support
prevents displacement of the end of the beams, but not its
rotation. The term roller supportrefers to a pin connection that
is free to move parallel to the axis of the beam; this type of
support suppresses only the transverse displacement.
Figure 4.1 Statically determinate beams
A cantilever beamis built into a rigid supportat one end, with the
other end being free,as shown in Fig.4.1(b). The builtin support
preventsdisplacementsas well asrotations of the end of the beam.
An overhanging beam, illustrated in Fig.4.1(c), is supported by a
pinand a roller support, with one or both ends of the beam
extending beyond thesupports.
The three types of beamsare statically determinatebecause the
support reactions can be found from the equilibrium equations.
(
)gg
(
)gg
(c) 2003 Brooks/Cole Publishing / Thomson Learning™
(c) 2003 Brooks/Cole Publishing / Thomson Learning™
Aconcentrated load, such as Pin Fig. 4.1(a). In contrast a
distributed loadis applied over a finite area. If the distributed
loadacts on a very narrow area, the load may be approximated by
a lineload.
The intensity wof this loading is expressed as force per unit
length (lb/ft, N/m, etc.) The load distribution may be uniform, as
shown in Fig.4.1(b), orit may vary with distancealong the beam,
as in Fig.4.1(c).
The weight of the beamis an example of distributed loading, but
its magnitude is usuallysmallcompared to the loads applied to
the beam.
(
)gg
(
)gg
(c) 2003 Brooks/Cole Publishing / Thomson Learning™
(c) 2003 Brooks/Cole Publishing / Thomson Learning™
Figure 4.2 Statically indeterminate beams
Figure 4.2 shows other types of beams. These beams are
oversupportedin the sense that each beam has at least one more
reaction than is necessary for support. Such beams arestatically
indeterminate; the presence of these redundant supports
requires the use of additional equationsobtained by considering
the deformation of the beam. The analysis of statically
indeterminate beams will be discussed in Chapter 7.
4.3 ShearMoment Equations and ShearMoment Diagrams
The determination of the internal force system acting at agiven
section of a beam: draw a freebody diagramthat expose these
forces and then compute the forces using equilibrium equations.
The goal of the beam analysis －determine the shear force Vand
the bending moment Matevery cross section of the beam.
To derive the expressionsfor Vand Min terms of the distance x
measured along the beam. By plotting these expressions to scale,
obtain theshear force and bending momentdiagramsfor the
beam.
The shear force and bending moment diagrams are convenient
visual references to the internal forces in a beam; in particular,
they identify the maximum valuesof Vand M.
a.Sign conventions
(
)k/lblihi/hi
(
)k/lblihi/hi
Figure 4.3 Sign conventions for external loads; shear force,
and bending moment.
b. Procedure for determining shear force and bending moment
diagrams
٠Compute the support reactionsfrom the freebody diagram
(FBD) of the entire beam.
٠Divide the beam into segmentso that the loading within each
segment iscontinuous. Thus, the endpointsof the segments are
discontinuities of loading, including concentrated loadsand
couples.
Perform the following steps for each segment of the beam:
Introduce an imaginary cutting planewithin the segment, located
at a distance x from the left end of the beam, that cuts the beam
into two parts.
Draw a FBD for the part of the beam lying either to the left or to
the right of the cutting plane, whichever is more convenient. At
the cut section, show V and M acting in their positive directions.
Determine the expressions for Vand Mfrom the equilibrium
equationsobtainable from the FBD. These expressions, which
are usually functions of x, are the shear force and bending
moment equations for the segment.
Plot the expressions for V and M for the segment. It is visually
desirable to draw the Vdiagrambelowthe FBD of the entire
beam, and then draw the MdiagrambelowtheVdiagram.
The bending moment and shear force diagrams of the beam are
composites of the Vand Mdiagrams of the segments. These
diagrams are usually discontinuous, or have discontinuous
slopes. At the endpointsof the segments due to discontinuities
in loading.
Sample Problem4.1
The simply supported beam in Fig. (a)carries two concentrated
loads. (1) Derive the expressions for the shear force and the bending
moment for each segment of the beam. (2) Sketch the shear force
and bending moment diagrams. Neglect the weight of the beam.
Note that the support reactions at Aand Dhave been computed and
are shown in Fig. (a).
Solution
Part 1
The determination of
the expressions for V
and Mfor each of the
three beam segments
(AB,BC, and CD) is
explained below.
SegmentAB (0＜x＜2 m)
ΣFy
=0 +↑18V= 0
V= +18 kNAnswer
ΣME
= 0 + 18x+ M= 0
M= +18x kN∙m Answer
SegmentAB (2＜x＜5 m)
ΣFy
=0 +↑1814V= 0
V= +1814 = +4 kNAnswer
ΣME
= 0 + 18x + 14(x2) + M= 0
M= +18x14(x2) = 4x+28 kN∙m Answer
Segment CD (5 m＜x＜7 m)
ΣFy
=0 +↑1814—28V= 0
V= +181428 = 24 kNAnswer
ΣMG
= 0 + 18x+ 14(x2)+28(x5)+M= 0
M= +18x14(x2) –(x5) = 24x+168 kN∙m Answer
Part 2
The Vdiagram reveals that
the largestshear forcein
the beam is 24 kN:
segment CD
The Mdiagram reveals
that the maximumbending
momentis +48 kN∙m: the
28kN load at C.
Note that at each
concentrated force the V
diagram “jumps”by an
amount equal tothe force.
There is a discontinuity in
the slopeof theMdiagram
at each concentrated force.
Sample problem4.2
The simply supported beam in Fig. (a) is loaded by theclockwise
couple C0
at B. (1) Derive the shear and bending moment
equations. And (2) draw the shear force and bending moment
diagrams. Neglect the weight of the beam. The support reactions A
and C have been computed, and their values are shown in Fig. (a).
Solution
Part 1
Due to the presence
of the coupleC0,
We
must analyze
segments ABand BC
separately.
Segment AB (0＜x＜3
L
/4)
Answer
Answer
Segment BC (3L/4＜x＜L)
Answer
Answer
00
0
=−−↑+=
∑
V
L
C
F
y
L
C
V
0
−=
∑
+=0
D
M
x
L
C
M
0
−=
0
0
=+Mx
L
C
00
0
=−−↑+=
∑
V
L
C
F
y
L
C
V
0
−=
∑
+=0
E
M
0
0
0
=+−MCx
L
C
0
0
Cx
L
C
M+−=
Part 2
From the Vdiagram,
the shear force is the
same for all cross
sections of the
beam. The M
diagram shows jump
of magnitudeC0
at
the point of
application of the
couple.
Sample Problem 4.3
The cantilever beam in
Fig.(a) carries a triangular
load. The intensity of
which varies from zero at
the left end to 360 lb/ft at
the right end. In addition, a
1000lb upward vertical
load acts at the free end of
the beam. (1) Derive the
shear force and bending
moment equations. And
(2) draw the shear force
and bending moment
diagrams. Neglect the
weight of the beam.
Solution
Note that the triangular load has been
replaced by is resultant, which is the
force 0.5 (12) (360) = 2160 lb (area
under the loading diagram) acting at
the centroidof the loading diagram.
Part 1
ΣFy
= 0 +↑1000 15x2
V= 0
V= 1000 15x2
lb Answer
ΣMC
= 0 + 1000x+ 15x2
(x/3) + M= 0
M = 1000x

5x3 lb∙ft Answer
Because the loading
is continuous, the
beam does nothave
to be divided into
segment.
w/x= 360/12, or
w= 30xlb/ft.
Part 2
The location of the
section where the
shear force is zerois
found from
V = 1000－15x2
= 0
x= 8.165 ft
x= 8.165 ft .
the maximumbending
moment is
Mmax
= 1000(8.165) 
5(8.165)3
= 5443 lb∙ft
0151000
2
=−=x
dx
dM
4.4 Area Methodfor Drawing ShearMoment Diagrams
Useful relationshipsbetween the loading, shear force, and
bending momentcan be derived from the equilibrium
equations.
These relationships enable us toplot the shear force diagram
directly from the load diagram, and then construct the bending
momentdiagramfrom the shear forcediagram. This technique,
called the area method, allows us to draw the shear force and
bending moment diagrams withouthaving to derive the
equations forVand M.
First consider beam subjected to distributed loadingand then
discuss concentrated forces and couples.
4.4 Figures (a) Simply
supported beam
carrying distributed
loading; (b) free
body diagram of an
infinitesimal beam
segment.
(c) 2003 Brooks/Cole Publishing / Thomson Learning™
(c) 2003 Brooks/Cole Publishing / Thomson Learning™
a.Distributed loading
Consider the beam in Fig. 4.4 (a) that is
subjected to a the distributed load w (x)
is assumed to be a continuous function.
The freebody diagram of an
infinitesimal element of the beam,
loaded at the distance xfrom the left
end, is shown in Fig. 4.4 (b)
The force equation of equilibrium is
ΣFy
= 0 +↑
V
－
wd
x－(
V
+
d V
) = 0
From which we get
(4.1)
The moment equation of equilibrium yields
ΣM0
= 0 + M–Vdx+ (M+dM) + wdx(dx/2) = 0
After canceling M and dividing by dx, we get
(4.2)
Equations (4.1) and (4.2) are called the differential equations of
equilibriumfor beams.
dx
dV
w−=
0
2
=++−
wdx
dx
dM
V
dx
dM
V=
The following five theoremsrelatingthe load, the shear force,
and the bending moment diagramsfollow from these
equations.
1.The load intensity wat any section of a beam is equal tothe
negativeof the slopeof the shear force diagram at the section.
Proof
－
follows directly from Eq. (4.1).
2. Theshearforce Vat any section is equal tothe slopeof the
bending moment diagram at that section.
Proof
－
follows directly from Eq. (4.2).
dx
dV
w−=
dx
dM
V=
3. The differencebetween the
shear forcesat two sectionsof a
beam is equal tothe negative of
the area under the load diagram
between those two sections.
Proof
－
integrating Eq. (4.1)
between sectionsAandBin
Fig. 4.5, we obtain
∫∫
−=−=
xB
xA
xB
xA
AB
wdxVVdx
dx
dV
VB－VA
=－area ofwdiagram]
A
B
VB
= VA－area of wdiagram] A
B
(4.3)
Note that the signs in Eq. (4.3) are correct only if xB＞xA.
dx
dV
w−=
4.The differencebetween the
bending momentsat two sections
of a beam is equal tothe area of
the shear force diagram between
these two sections.
Proof
－
integratingEq. (4.2)
between sectionsAandBin ( see
Fig. 4.5),
MB－MA
= area of Vdiagram]A
B
Q.E.D
MB=MA
+ area of Vdiagram]A
B
(4.4)
The signs in Eq. (4.4) are correct only if xB＞xA.
∫∫
=−=
XB
XA
XB
XA
AB
VdxMMdx
dx
dM
dx
dM
V=
5.If the loadWdiagram is a polynomial of degreen, then shear
force V diagram is polynomial of degree ( n+1), and the
bending moment Mdiagram is polynomial of degree ( n+2).
Proof
－
followings directly from the integration of Eqs. (4.1)
and (4.2).
Consider the beam segment shown in Fig. 4.6 (a), which is 2 m
long and is subjected to a uniformly distributed load w= 300 N
/m. Figure 4.6 (a) shows the shear force and the bending
moment at the left end are VA= +1000 N and MA
= +3000 N∙m.
dx
dM
V=
dx
dV
w−=
Figure 4.6 (a) Freebody diagram of a beam segment carrying
uniform loading;
Figure 4.6(b) constructing shear force and bending moment
diagrams for the beam segment.
Figure 4.7 Freebody diagram
of an infinitesimal
beam element carrying
a concentrated force PA
and a concentrated
couple CA.
b.Concentrated forces and couples.
The force equilibrium equation
ΣFy
= 0 + ↑VA

PA VA
+ = 0
Equation (4.5) indicates that apositive concentrated force
causes a negative jump discontinuityin the shear force diagram
at A (a concentrated coupledoes not affect the shear force
diagram).
V
A
+ = V
A
P
A
(4.5)
The moment equilibrium equation yields
ΣMA = 0 +
0
22
=−−−−
−+−+
dx
V
dx
VCMM
AAAAA
AAA
CMM+=
−+
Thus, a positive concentrated
couplecauses a positive jump
in the bending moment
diagram.
Figure 4.7 Freebody diagram of an infinitesimal beam
element carrying a concentrated force PA
and a
concentrated couple C
A.
c.Summary
The area method is useful only if the area under the load and shear
force diagrams can be easily computed.
(4.1)
(4.2)
(4.3)
(4.4)
(4.5)
(4.6)
dx
dM
V=
dx
dV
w−=
AAA
PVV−=
−+
AAA
CMM+=
−+
VB
= V
A
－area of wdiagram]A
B
M
B = M
A
+ area of wdiagram]
A
B
Procedure for the Area Method
Compute the support reactions forcethe freebody diagrams
(FBD) of the entire beam.
Draw the load diagram of the beam (which is essentially a
FBD) showing the values of the loads, including the support
reactions.Use the sign conventions in Fig. 4.3 to determine
the correct sign of each load.
Working from left to right, construct the Vand Mdiagram
for each segment of the beam using Eqs. (4.1)(4.6).
When reach the right end of the beam, checkto see whether
the computed values of Vand Mare consistent with the end
conditions. If they are not, you made an error in the
computations.
Sample Problem4.4
The simply supported beam in Fig. (a) supports 30kN concentrated
force at Band a 40kN•m couple at D.Sketch the shear force and
beading moment diagrams by the area method. Neglect the weight
of the beam.
Solution
Load Diagram
The load diagram for the beam is shown in Fig. (b). The reactions
at Aand Eare found from equilibrium analysis. Indicating its sign
as established by the sign conventions in Fig. 4.3.
Shear Force Diagram
There are concentrated forces at
A,B, noting that VA
= 0 because
no load is applied to the left of A
VB

= VA

area of wdiagram]A
B
=140 = 14 kN
Because w=dV/dx= 0 between
Aand B,
(
)
kNPVV
BBB
163014−=+−=−=
−+
(
)
01616=−−−=−=
−+
EEE
RVV
Check
！
(
)
(
)
kNRVV
AAA
14140+=−−=−=
−+
P
lot point.
○
a
P
lot point.
○
b
Connect
○
a湤n
○
戠with
VE

= VB
+
area of wdiagram]B
E = 160 = 16 kN
P
lot point.
○
c
P
lot point.
○
d
Connect
○
c
and
○
d
with a horizontal line
Because w=－dV/dx= 0 between Band E
a horizontal straight line
Bending Moment Diagram
The applied couple is cause a
jump in the bending moment
diagram at D.
The areas are either positive
or negative, depending on the
sign of the shear force in Fig.
(c).M
A
= 0 (there is no couple
applied atA). point (e)
MB
= MA
+ area of Vdiagram]A
B
= 0 + (+56) = 56 kN∙m , point (f)
The shear force bewteenA and B is constant and positive. The
slope of the Mdiagram between these two sections is also constant
and positive. ( recall that V=dM/dx= 0 ), connect (e) and (f) with
straight line.
MD

= MB
+ area of Vdiagram]B
D
= 56 + (48) = 8 kN∙m , point (g)
The slope of the V diagram
between B and D is negative and
constant, the Mdiagram has a
constant, negative slope in this
segment, ), connect (f) and (g)
with straight line.
MD
+
= MD

+ CD
= 8 + (+40) = 48 kN∙m
Point (h), note thatME
= 0 (these
is no couple applied atE).
ME = M
D
+
+ area of Vdiagram]
D
E
= 48 + (48) = 0 Check
！
The shear force between D and E
is negative and constant, which
means that the slope of the M 
diagram for this segment is also
constant and negative, connect (h)
and (i) with straight line.
Sample Problem4.5
The overhanging beam in Fig. (a) carries two uniformly distributed
loads and a concentrated load Using the area method. Draw the
shear force and bending moment diagrams for the beam.
Load Diagram
The load
diagram for the
beam is given
in Fig. (b)
Shear Force
Diagram
The steps
required to
construct the
shear force
diagram in Fig.
(c) are now
detailed.
Bending
Moment
Diagram
The slope of
theM
diagram is
discontinuo
us atj and
m.
Σχόλια 0
Συνδεθείτε για να κοινοποιήσετε σχόλιο